Mathematics
WAEC 1999
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
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A.
\(\frac{\sqrt{3}+1}{2}\)
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B.
\(\frac{2}{\sqrt{3}+1}\)
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C.
\(\frac{\sqrt{3}-1}{2}\)
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D.
\(\frac{2}{\sqrt{3}-1}\)
Correct Answer: Option C
Explanation

\(\cos x = \frac{\sqrt{3}}{2}\)
\(\sin x = \frac{1}{2}\)
\(\cos x - \sin x = \frac{\sqrt{3} - 1}{2}\)
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